Proving Divisibility by 9 using Induction

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To prove that 4^n + 15n - 1 is divisible by 9 for all n > 0, the initial step involves verifying that f(1) is divisible by 9. The inductive hypothesis assumes that f(n) is divisible by 9, leading to the need to show that f(n+1) is also divisible by 9. The discussion highlights a struggle with the algebraic manipulation required for the inductive step. Ultimately, the original poster successfully finds the solution after receiving guidance. The thread emphasizes the importance of mathematical induction in proving divisibility.
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Homework Statement



prove that for all n>0 , 4^n + 15n - 1 is divisible by 9/multiple of 9

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The Attempt at a Solution


need to show: 4^(n+1) + 15(n+1) + 14/9 = a*k a and k are integers
assumption to inductive step: (4^n + 15n - 1)/9 = k --> 4^n + 15n - 1 = 9k -->
4^n+1 + 60n - 4 = 36k.. now what? I tried 4^n+1 -4 = 6(6k - 10n) but ultimately , I am stuck

please help! tank you
 
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holezch said:
need to show: 4^(n+1) + 15n + 14/9 = a*k a and k are integers
Where did you get this?

Let f(n) = 4^n + 15n - 1. First, you need to show that f(1) is divisible by 9. Then, you need to show that f(n+1) is divisible by 9 if f(n) is.
 


sorry , I had a typo. thanks for the reply but I got the answer!
 

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